Let $f(x, y, z) = z - 3x^2 + 2y^2$ and $g(t) = (2t, \sqrt{2t}, t)$. $h(t) = f(g(t))$ $h'(2) = $
Solution: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(2) = \nabla f(g(2)) \cdot g'(2)$. $\begin{aligned} &g(2) = (4, 2, 2) \\ \\ &g'(2) = \left( 2, \dfrac{2}{2\sqrt{2t}}, 1 \right) = \left( 2, \dfrac{1}{2}, 1 \right) \\ \\ &\nabla f = \left( -6x, 4y, 1 \right) \\ \\ &\nabla f(g(2)) = \nabla f(4, 2, 2) = \left( -24, 8, 1 \right) \end{aligned}$ Substituting: $\begin{aligned} h'(2) &= (-24, 8, 1) \cdot \left( 2, \dfrac{1}{2}, 1 \right) \\ \\ &= -48 + 4 + 1 \\ \\ &= -43 \end{aligned}$ Answer $h'(2) = -43$